Suppose $G({\mathbb Z})$ is a higher rank non-cocompact arithmetic group (e.g. $SL_n({\mathbb Z})$ with $n\geq 3$, or $Sp_{2g}({\mathbb Z})$ with $g\geq 2$). I have seen a result (http://arxiv.org/abs/math/0409345) which says that every finite index subgroup $\Gamma $ of $G({\mathbb Z})$ contains a smaller finite index subgroup generated by three elements.

Does anyone know ANY example of $G({\mathbb Z})$, where three can be replaced by two? I believe Alan Reid has some result in this direction.

[Edit] That 2 should suffice is a conjecture, attributed to Alex Lubotzky. That $3$ DO suffice for non-uniform higher rank lattices in the result mentioned in the link. What I am asking is just ONE example where 2 generators suffice.

Just a remark: for arithmetic subgroups of $PSL(2,C)$, it is conjectured there are finitely many 2-generated arithmetic groups. This is known to be a consequence of Lehmer's conjecture.
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Ian AgolMar 2 '13 at 16:57

Thank you. This is most interesting.
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VenkataramanaMar 2 '13 at 17:03

1 Answer
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The result to which you refer is not a result but a conjecture of A. Lubotzky. Long and Reid have constructed some examples. -- the relevant preprints can be found on Alan Reid's web page. I assume that Lubotzky's conjecture is about three and not two generators because he did not want to be too ambitious -- nobody knows anything concrete, to the best of my knowledge.

Igor, Thanks very much for the link to Alan Reid's paper. The result that I was thinking about was one by Ritumoni Sarma, where he proved the three generator (non-cocompact higher rank lattice) property. However, no one to whom I have talked (admittedly a small number!) could give me an example where two suffice. For co-compact, you are right: nobody (to whom I have talked) knows any result.
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VenkataramanaNov 19 '12 at 6:08

Actually, I didn't know about Sarma's result (actually, it's Sarma and Venkataramana)! Thanks for the reference!
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Igor RivinNov 19 '12 at 13:01